# proof of matrix inverse calculation by Gaussian elimination

Let $A$ be an invertible matrix, and ${A}^{-1}$ its inverse^{}, whose
columns are ${A}_{1}^{-1},\mathrm{\cdots},{A}_{n}^{-1}$.
Then, by definition of matrix inverse, $A{A}^{-1}={I}_{n}$.
But this implies
$A{A}_{1}^{-1}={e}_{1},\mathrm{\dots},A{A}_{n}^{-1}={e}_{n}$,
with ${e}_{1},\mathrm{\cdots},{e}_{n}$ being the first,$\mathrm{\dots}$,$n$-th column of ${I}_{n}$
respectively.

$A$ being non singular (or invertible), for all $k\le n$, $A{A}_{k}^{-1}={e}_{k}$ has a solution for ${A}_{k}^{-1}$, which can
be found by Gaussian elimination^{} of $[A\mid {e}_{k}]$.

The only part that changes between the augmented matrices constructed is the last column, and these last columns, once the Gaussian elimination has been performed, correspond to the columns of ${A}^{-1}$. Because of this, the steps we need to take for the Gaussian elimination are the same for each augmented matrix.

Therefore, we can solve the matrix equation by performing Gaussian elimination on $[A\mid {e}_{1}\mathrm{\cdots}{e}_{n}]$, or $[A\mid {I}_{n}]$.

Title | proof of matrix inverse calculation by Gaussian elimination |
---|---|

Canonical name | ProofOfMatrixInverseCalculationByGaussianElimination |

Date of creation | 2013-03-22 14:15:10 |

Last modified on | 2013-03-22 14:15:10 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 9 |

Author | rspuzio (6075) |

Entry type | Definition |

Classification | msc 15A09 |